Verified answer

The height of a projectile at any time t is given by y(t) = h + Uyt - 1/2 gt^2; where h = 1.5 m and Uy = ? is the vertical launch speed you're looking for.

Find Vy = 0 = Uy - gt, where Vy reaches zero just as the ball hits the ceiling. Thus Uy = gt and we have y(t) = 10 = 1.5 + gt^2 - 1/2 gt^2 and 8.5 = 1/2 gt^2 so that t = 17/9.8 = 1.73 seconds rise time. And Uy = 9.8*1.73 = 16.95 mps ANS.

When dropping from H = 10 m, the impact speed Vy = sqrt(2*9.8*10) = 14 mps. ANS.

If we take the acceleration due to gravity as 10m/s^2 it will be easier to see.

This means that if you throw something straight up, then, every second, its velocity will fall by 10m/s. So, if you throw it up with a velocity of 10m/s, it will come to a stop after 1 second, and fall back to the ground.

So, to answer your first part, the minimum velocity required to attain a height of 10 meters is 10 m/s.

For the second part, if you release the ball from a height of 1.5 m, its vertical height is given by

V(h) = (1/2) g t^2 m

where g = acceleration due to gravity = 10 m/s^2

So,

1.5 = 0.5 * 10 * t^2 = 5 t^2

t^2 = 1.5 / 5 = 0.3

t = √0.3 = 0.55 s to two decimal places.

If you require better accuracy, just use 9.81 m/s^2 instead of 10 m/s^2 for the acceleration due to gravity.